Question

A reaction A-> B has the following time dependence for the concentration of [A] vs time. For t=(0 s, 5 s, 10 s, 15 s, 25 s) the concentration of [A]=(30.00 M, 15.42 M, 10.38 M, 7.82 M, 5.24 M). The initial concentration of [A] is the value at t=0 s. (A)Calculate the values of the rate constant k assuming that the reaction is first order. - for all values k

B)calculate the value of k if the reaction is Second Order

C)What is the concentration of [A] at time t=72 s?

Answer 1

Asssuming order of the reaction is first order

First order rate equation

Rate =k[A]¹

Rate constant

ln(15.42) = -k(5) + ln(30)

2.74 = -5k + 3.40

-0.66 = -5k

Rate constant at 5s, k = 0.132 s^-1

Rate constant at 10s =

ln(10.38) = -k(10) + 3.40

2.34 = -10k + 3.40

Rate constant at 10s = k = 0.106^-1

Rate constnat 15s =

ln7.82 =-k(15) + 3.40

k= 0.089 s^-1

Rate constant at 15s = k= 0.089 s^-1

Rate constant at 25s =

ln(5.24) = -kt + 3.40

1.656 = -k(25) + 3.40

k = 0.0697 s^-1

Rate constant at 25 s = 0.0697 s^-1

The K values are not consistent

So, from these k values it is clear that the order of the reaction is not first order

b) If reaction is second order

now second order rate constant =

to check let us

Plot the graph between 1/[A] vs t

slope = rate costant, k = 0.0063 M^-1. s^-1

C) Now calculating concentration of [A] at 72 s =

this reaction second order so using second order rate constant equation

[A] = 2.05 M

Concentration of [A] = 2.05 M